Question

In fig. BD || CA, E is mid-point of CA and $BD=\frac{1}{2}$AC.Then

• A. $ar\left(ABC\right)=\frac{1}{2}ar\left(DBC\right)$
• B. $ar\left(ABC\right)=ar\left(DBC\right)$
• C. $ar\left(ABC\right)=2ar\left(DBC\right)$
• D. $ar\left(ABC\right)=\frac{3}{2}ar\left(DBC\right)$

Answer 1

The correct option is C $ar\left(ABC\right)=2ar\left(DBC\right)$

Here, BCED is a parallelogram, since
BD = CE and BD||CE
ar(DBC) = ar(EBC) ……(i)
In $\Delta ABC$, BE is the median,
So, $ar\left(EBC\right)=\frac{1}{2}ar\left(ABC\right)$
Now, ar(ABC) = ar(EBC)+ar(ABE)
Also, ar(ABC) = 2 ar(EBC), therefore,
ar(ABC) = 2ar(DBC).